Class 12 Physics Alternating Current Power associated with Inductor

Power associated with Inductor

  • Voltage and current associated with the inductor V = Vm sinωt  is  I = Im sin(ωt –(∏/2)) respectively.
  • Instantaneous power p = VI = Vm Im sinωt sin(ωt –(∏/2))
  • = - Vm Im sinωt cos ωt  ( using  formula (sin(A –(∏/2)) = – cosA))
  • After simplifying we get,
  • p = - ((Vm Im sin2ωt)/(2))
  • Average Power PL = (1/T) 0T∫ p dt
    • =(-1/T) 0T∫ (Vm Im/2) sin2 ωt dt
    • = - ( Vm Im/2T) 0T∫ sin2 ωt dt
    • After simplifying  we will get ,
  • PL  =  0.
  • Therefore average power supplied to an inductor over one complete cycle is 0.

Problem:- A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.


The inductive reactance,

XL = 2πν L = (2x3.14x50x25x10-3) W

= 7.85Ω

The rms current in the circuit is:- I =(V/XL) =(220V)/(7.85Ω) =28A


Problem:-A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V,

50 Hz ac supply. (a) What is the maximum current in the coil? (b) What is the time lag between the voltage maximum and the current maximum?


Inductance of the inductor, L = 0.50 H

Resistance of the resistor, R = 100 Ω

Potential of the supply voltage, V = 240 V

Frequency of the supply, ν = 50 Hz

(a) Peak voltage is given as: V0=√2V

=√2 x 240 =339.41V

Angular frequency of the supply, ω = 2 πν

= 2π × 50 = 100 π rad/s

Maximum current in the circuit is given as:

I0= (V0)/( √(R)2 + ω2L2)

=(339.41)/((100)2+(100π)2+(0.50)2) = 1.82A

(b) Equation for voltage is given as:

V = V0 cos ωt

Equation for current is given as:

I = I0 cos (ωt − Φ)


Φ = Phase difference between voltage and current At time, t = 0.

V = V0(voltage is maximum)

For ωt − Φ = 0 i.e., at time ,

I = I0 (current is maximum)

Hence, the time lag between maximum voltage and maximum current is .( Φ/ω)

Now, phase angle Φ is given by the relation,

tan Φ = (ωL)/(R)

= (2 π x 50x 0.5)/(100)


Φ = 57.50 = (57.5 π)/(180)rad

ωt = (57.5 π)/(180)

t =(57.5)/(180 x 2 π x 50)

=3.19x10-3 s


Hence, the time lag between maximum voltage and maximum current is 3.2 ms.

Problem:- Obtain the answers (a) to (b) in previous problem if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?


Inductance of the inductor, L = 0.5 Hz

Resistance of the resistor, R = 100 Ω

Potential of the supply voltages, V = 240 V

Frequency of the supply, ν = 10 kHz = 104Hz

Angular frequency, ω = 2πν= 2π × 104rad/s

  • Peak voltage, V0=√2V = 240√2 V

Maximum current , I0= (V0)/( √(R)2 + ω2L2)

=(240√2) / ((100)2 + (2 π x 104)2 x(0.50)2)

=1.1 x 10-2 A

 (b) For phase difference Φ, we have the relation:

tan Φ = (ωL)/(R)

= (2 π x 104 x 0.5)/(100) =100 π

Φ =89.820 = (89.82 π)/(180) rad

ω t = (89.82 π)/(180)

t= (89.82 π)/(180x2 πx104)

=25 μ s

It can be observed that I0 is very small in this case. Hence, at high

frequencies, the inductor amounts to an open circuit.

In a dc circuit, after a steady state is achieved, ω = 0. Hence, inductor L behaves like a pure conducting object.

(Magnetisation and demagnetisation of a magnet)

  • Current i through the coil entering at point A increase from zero to a maximum value. Flux lines are set up i.e., the core gets magnetised. With the polarity shown voltage and current are both positive. So their product p is positive.


  • Current in the coil is still positive but is decreasing. The core gets demagnetised and the net flux becomes zero at the end of a half cycle. The voltage v is negative (since di/dt is negative). The product of voltage and current is negative, and ENERGY IS BEING RETURNED TO SOURCE.

(One complete cycle of voltage/current . It is clear  that the current lags  the voltage.)

  • Current i becomes negative i.e., it enters at B and comes out of A. Since the direction of current has changed, the polarity of the magnet changes. The current and voltage are both negative. So their product p is positive. ENERGY IS ABSORBED.

  • Current i decreases and reaches its zero value at 4 when core is demagnetised and flux is zero. The voltage is positive but the current is negative. The power is, therefore, negative. ENERGY ABSORBED DURING THE ¼ CYCLE 2-3 IS RETURNED TO THE SOURCE.

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